# The bisector of the exterior angle ∠A of Δ ABC intersects side BC produced at D. Prove that .

Given:
ABC is a triangle; AD is the exterior bisector of ∠A and meets BC produced at D; BA is produced to F.
To prove:
Construction: Draw CE||DA to meet AB at E.
Proof: In Δ ABC, CE || AD cut by AC.
Similarly CE || AD cut by AB
∴ ∠ACE = ∠AEC
AC = AE ( by isosceles Δ theorem)
Now in Δ BAD, CE || DA

(Basic proportionality theorem):If a line is drawn parallel to one side of a triangle intersecting other two sides,
then it divides the two sides in the same ratio. '

By BPT,

But AC = AE (proved above)

or (proved).

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